The decomposition of $N_2O_5$ in $CCl_4$ at 318K has been studied by monitoring the concentration of $N_2O_5$ in the solution. Initially, the concentration of $N_2O_5$ is 2.33 mol/L and after 184 minutes, it is reduced to 2.08 mol/ L. The reaction takes place according to the equation
$2\text{N}_2\text{O}_5(\text{g})\rightarrow4\text{NO}_2(\text{g})+\text{O}_2(\text{g})$
The average rate of this reaction in different units will be:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $4.07\times10^{-2}\text{ mol L}^{-1}\text{ h}^{-1},6.79\times10^{-4}\text{ mol L}^{-1}\text{ min}^{-1},1.13\times10^{-5}\text{ mol L}^{-1}\text{ s}^{-1}$
Average rate: The average rate of a chemical reaction is the change in concentration of a reactant or product over a specific time interval
Rate of disappearance of $R=-\frac{\Delta[R]}{\Delta t}$
Given in question:
$\text{Initial }[\mathrm{N_2O}_5]=2.33\,\text{mol L}^{-1}$
$\text{Final }[\mathrm{N_2O}_5]=2.08\,\text{mol L}^{-1}\quad(\text{Disappearance of reactant})$
$\text{Taking stoichiometric coefficient into consideration:}$
$\text{Average Rate}=-\frac12\left[\frac{\Delta[\text{N}_2\text{O}_5]}{\Delta t}\right]=-\frac12\left[\frac{(2.08-2.33)\text{ mol L}^{-1}}{184\text{ min}}\right]$
$\frac{1}{184\,\text{min}}\left(2.08-2.33\right)\,\text{mol L}^{-1}$
$=6.79\times10^{-4}\,\text{mol L}^{-1}\text{min}^{-1}$
$=\left(6.79\times10^{-4}\,\text{mol L}^{-1}\text{min}^{-1}\right)\times\frac{60\,\text{min}}{1\,\text{h}}$
$=4.07\times10^{-2}\,\text{mol L}^{-1}\text{h}^{-1}$
$6.79\times10^{-4}\,\text{mol L}^{-1}\text{min}^{-1}\times\frac{1\,\text{min}}{60\,\text{s}}$
$=1.13\times10^{-5}\,\text{mol L}^{-1}\text{s}^{-1}$